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### Chapt07

1. 1. CHAPTER 7 POISSON'S AND LAPLACE'S EQUATIONS A study of the previous chapter shows that several of the analogies used to obtain experimental field maps involved demonstrating that the analogous quan- tity satisfies Laplace's equation. This is true for small deflections of an elastic membrane, and we might have proved the current analogy by showing that the direct-current density in a conducting medium also satisfies Laplace's equation. It appears that this is a fundamental equation in more than one field of science, and, perhaps without knowing it, we have spent the last chapter obtaining solu- tions for Laplace's equation by experimental, graphical, and numerical methods. Now we are ready to obtain this equation formally and discuss several methods by which it may be solved analytically. It may seem that this material properly belongs before that of the previous chapter; as long as we are solving one equation by so many methods, would it not be fitting to see the equation first? The disadvantage of this more logical order lies in the fact that solving Laplace's equation is an exercise in mathe- matics, and unless we have the physical problem well in mind, we may easily miss the physical significance of what we are doing. A rough curvilinear map can tell us much about a field and then may be used later to check our mathematical solutions for gross errors or to indicate certain peculiar regions in the field which require special treatment. With this explanation let us finally obtain the equations of Laplace and Poisson. 195 | | | | v v e-Text Main Menu Textbook Table of Contents
2. 2. 7.1 POISSON'S AND LAPLACE'S EQUATIONS Obtaining Poisson's equation is exceedingly simple, for from the point form of Gauss's law, r Á D  v 1 the definition of D, D  E 2 and the gradient relationship, E  ÀrV 3 by substitution we have r Á D  r Á E  Àr Á rV  v or r Á rV  À v 4 for a homogeneous region in which is constant. Equation (4) is Poisson's equation, but the ``double r'' operation must be interpreted and expanded, at least in cartesian coordinates, before the equation can be useful. In cartesian coordinates, r Á A  @Ax @x  @Ay @y  @Az @z rV  @V @x ax  @V @y ay  @V @z az and therefore r Á rV  @ @x @V @x  @ @y @V @y  @ @z @V @z  @2 V @x2  @2 V @y2  @2 V @z2 5 Usually the operation r Á r is abbreviated r2 (and pronounced ``del squared''), a good reminder of the second-order partial derivatives appearing in (5), and we have 196 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
3. 3. r2 V  @2 V @x2  @2 V @y2  @2 V @z2  À v 6 in cartesian coordinates. If v  0, indicating zero volume charge density, but allowing point charges, line charge, and surface charge density to exist at singular locations as sources of the field, then r2 V  0 7 which is Laplace's equation. The r2 operation is called the Laplacian of V. In cartesian coordinates Laplace's equation is r2 V  @2 V @x2  @2 V @y2  @2 V @z2  0 cartesian 8 and the form of r2 V in cylindrical and spherical coordinates may be obtained by using the expressions for the divergence and gradient already obtained in those coordinate systems. For reference, the Laplacian in cylindrical coordinates is r2 V  1 @ @ @V @  1 2 @2 V @2  @2 V @z2 cylindrical 9 and in spherical coordinates is r2 V  1 r2 @ @r r2 @V @r  1 r2 sin @ @ sin @V @  1 r2 sin2 @2 V @2 spherical 10 These equations may be expanded by taking the indicated partial derivatives, but it is usually more helpful to have them in the forms given above; furthermore, it is much easier to expand them later if necessary than it is to put the broken pieces back together again. Laplace's equation is all-embracing, for, applying as it does wherever volume charge density is zero, it states that every conceivable configuration of electrodes or conductors produces a field for which r2 V  0. All these fields are different, with different potential values and different spatial rates of change, yet for each of them r2 V  0. Since every field (if v  0 satisfies Laplace's equa- tion, how can we expect to reverse the procedure and use Laplace's equation to find one specific field in which we happen to have an interest? Obviously, more POISSON'S AND LAPLACE'S EQUATIONS 197 | | | | v v e-Text Main Menu Textbook Table of Contents
4. 4. information is required, and we shall find that we must solve Laplace's equation subject to certain boundary conditions. Every physical problem must contain at least one conducting boundary and usually contains two or more. The potentials on these boundaries are assigned values, perhaps V0, V1; . . . , or perhaps numerical values. These definite equi- potential surfaces will provide the boundary conditions for the type of problem to be solved in this chapter. In other types of problems, the boundary conditions take the form of specified values of E on an enclosing surface, or a mixture of known values of V and E: Before using Laplace's equation or Poisson's equation in several examples, we must pause to show that if our answer satisfies Laplace's equation and also satisfies the boundary conditions, then it is the only possible answer. It would be very distressing to work a problem by solving Laplace's equation with two different approved methods and then to obtain two different answers. We shall show that the two answers must be identical. D7.1. Calculate numerical values for V and v at point P in free space if: a V  4yz x2  1 , at P1; 2; 3; b V  52 cos 2, at P  3;  3 ; z  2; c V  2 cos r2 , at Pr  0:5;  458,  608: Ans. 12 V, À106:2 pC/m3 ; 22.5 V, 0; 4 V, À141:7 pC/m3 7.2 UNIQUENESS THEOREM Let us assume that we have two solutions of Laplace's equation, V1 and V2, both general functions of the coordinates used. Therefore r2 V1  0 and r2 V2  0 from which r2 V1 À V2  0 Each solution must also satisfy the boundary conditions, and if we repre- sent the given potential values on the boundaries by Vb, then the value of V1 on the boundary V1b and the value of V2 on the boundary V2b must both be identical to Vb; V1b  V2b  Vb or V1b À V2b  0 198 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
5. 5. In Sec. 4.8, Eq. (44), we made use of a vector identity, r Á VD Vr Á D  D Á rV which holds for any scalar V and any vector D. For the present application we shall select V1 À V2 as the scalar and rV1 À V2 as the vector, giving r Á V1 À V2rV1 À V2 V1 À V2r Á rV1 À V2  rV1 À V2 Á rV1 À V2 which we shall integrate throughout the volume enclosed by the boundary surfaces specified:  vol r Á V1 À V2rV1 À V2 dv  vol V1 À V2r Á rV1 À V2dv   vol rV1 À V22 dv 11 The divergence theorem allows us to replace the volume integral on the left side of the equation by the closed surface integral over the surface surrounding the volume. This surface consists of the boundaries already specified on which V1b  V2b, and therefore  vol r Á V1 À V2rV1 À V2 dv   S V1b À V2brV1b À V2b Á dS  0 One of the factors of the first integral on the right side of (11) is r Á rV1 À V2, or r2 V1 À V2, which is zero by hypothesis, and therefore that integral is zero. Hence the remaining volume integral must be zero:  vol rV1 À V22 dv  0 There are two reasons why an integral may be zero: either the integrand (the quantity under the integral sign) is everywhere zero, or the integrand is positive in some regions and negative in others, and the contributions cancel algebraically. In this case the first reason must hold because rV1 À V22 can- not be negative. Therefore rV1 À V22  0 and rV1 À V2  0 Finally, if the gradient of V1 À V2 is everywhere zero, then V1 À V2 cannot change with any coordinates and V1 À V2  constant If we can show that this constant is zero, we shall have accomplished our proof. The constant is easily evaluated by considering a point on the boundary. Here POISSON'S AND LAPLACE'S EQUATIONS 199 | | | | v v e-Text Main Menu Textbook Table of Contents
6. 6. V1 À V2  V1b À V2b  0, and we see that the constant is indeed zero, and there- fore V1  V2 giving two identical solutions. The uniqueness theorem also applies to Poisson's equation, for if r2 V1  Àv= and r2 V2  Àv=, then r2 V1 À V2  0 as before. Boundary conditions still require that V1b À V2b  0, and the proof is identical from this point. This constitutes the proof of the uniqueness theorem. Viewed as the answer to a question, ``How do two solutions of Laplace's or Poisson's equation com- pare if they both satisfy the same boundary conditions?'' the uniqueness theorem should please us by its ensurance that the answers are identical. Once we can find any method of solving Laplace's or Poisson's equation subject to given boundary conditions, we have solved our problem once and for all. No other method can ever give a different answer. D7.2. Consider the two potential fields V1  y and V2  y  ex sin y. a Is r2 V1  0? b Is r2 V2  0? c Is V1  0 at y  0? d Is V2  0 at y  0? e Is V1  at y  ? f  Is V2  at y  ? g Are V1 and V2 identical? h Why does the uniqueness theorem not apply? Ans. Yes; yes; yes; yes; yes; yes; no; boundary conditions not given for a closed surface 7.3 EXAMPLES OF THE SOLUTION OF LAPLACE'S EQUATION Several methods have been developed for solving the second-order partial differ- ential equation known as Laplace's equation. The first and simplest method is that of direct integration, and we shall use this technique to work several exam- ples in various coordinate systems in this section. In Sec. 7.5 one other method will be used on a more difficult problem. Additional methods, requiring a more advanced mathematical knowledge, are described in the references given at the end of the chapter. The method of direct integration is applicable only to problems which are ``one-dimensional,'' or in which the potential field is a function of only one of the three coordinates. Since we are working with only three coordinate systems, it might seem, then, that there are nine problems to be solved, but a little reflection will show that a field which varies only with x is fundamentally the same as a field which varies only with y. Rotating the physical problem a quarter turn is no change. Actually, there are only five problems to be solved, one in cartesian coordinates, two in cylindrical, and two in spherical. We shall enjoy life to the fullest by solving them all. 200 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
7. 7. hExample 7.1 Let us assume that V is a function only of x and worry later about which physical problem we are solving when we have a need for boundary conditions. Laplace's equa- tion reduces to @2 V @x2  0 and the partial derivative may be replaced by an ordinary derivative, since V is not a function of y or z, d2 V dx2  0 We integrate twice, obtaining dV dx  A and V  Ax  B 12 where A and B are constants of integration. Equation (12) contains two such constants, as we should expect for a second-order differential equation. These constants can be determined only from the boundary conditions. What boundary conditions should we supply? They are our choice, since no physical problem has yet been specified, with the exception of the original hypothesis that the potential varied only with x. We should now attempt to visualize such a field. Most of us probably already have the answer, but it may be obtained by exact methods. Since the field varies only with x and is not a function of y and z, then V is a constant if x is a constant or, in other words, the equipotential surfaces are described by setting x constant. These surfaces are parallel planes normal to the x axis. The field is thus that of a parallel-plate capacitor, and as soon as we specify the potential on any two planes, we may evaluate our constants of integration. To be very general, let V  V1 at x  x1 and V  V2 at x  x2. These values are then substituted into (12), giving V1  Ax1  B A  V1 À V2 x1 À x2 V2  Ax2  B B  V2x1 À V1x2 x1 À x2 and V  V1x À x2 À V2x À x1 x1 À x2 A simpler answer would have been obtained by choosing simpler boundary conditions. If we had fixed V  0 at x  0 and V  V0 at x  d, then A  V0 d B  0 POISSON'S AND LAPLACE'S EQUATIONS 201 | | | | v v e-Text Main Menu Textbook Table of Contents
8. 8. and V  V0x d 13 Suppose our primary aim is to find the capacitance of a parallel-plate capacitor. We have solved Laplace's equation, obtaining (12) with the two constants A and B. Should they be evaluated or left alone? Presumably we are not interested in the potential field itself, but only in the capacitance, and we may continue successfully with A and B or we may simplify the algebra by a little foresight. Capacitance is given by the ratio of charge to potential difference, so we may choose now the potential difference as V0, which is equivalent to one boundary condition, and then choose whatever second boundary condition seems to help the form of the equation the most. This is the essence of the second set of boundary conditions which produced (13). The potential difference was fixed as V0 by choosing the potential of one plate zero and the other V0; the location of these plates was made as simple as possible by letting V  0 at x  0: Using (13), then, we still need the total charge on either plate before the capacitance can be found. We should remember that when we first solved this capacitor problem in Chap. 5, the sheet of charge provided our starting point. We did not have to work very hard to find the charge, for all the fields were expressed in terms of it. The work then was spent in finding potential difference. Now the problem is reversed (and simplified). The necessary steps are these, after the choice of boundary conditions has been made: 1. Given V, use E  ÀrV to find E: 2. Use D  E to find D: 3. Evaluate D at either capacitor plate, D  DS  DNaN: 4. Recognize that S  DN: 5. Find Q by a surface integration over the capacitor plate, Q   S SdS: Here we have V  V0 x d E  À V0 d ax D  À V0 d ax 202 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
9. 9. DS  D x0  À V0 d ax aN  ax DN  À V0 d  S Q   S ÀV0 d dS  À V0S d and the capacitance is C  jQj V0  S d 14 We shall use this procedure several times in the examples to follow. hExample 7.2 Since no new problems are solved by choosing fields which vary only with y or with z in cartesian coordinates, we pass on to cylindrical coordinates for our next example. Variations with respect to z are again nothing new, and we next assume variation with respect to only. Laplace's equation becomes 1 @ @ @V @  0 or 1 d d dV d  0 Noting the in the denominator, we exclude  0 from our solution and then multiply by and integrate, dV d  A rearrange, and integrate again, V  A ln  B 15 The equipotential surfaces are given by  constant and are cylinders, and the problem is that of the coaxial capacitor or coaxial transmission line. We choose a potential difference of V0 by letting V  V0 at  a, V  0 at  b, b a, and obtain V  V0 lnb= lnb=a 16 POISSON'S AND LAPLACE'S EQUATIONS 203 | | | | v v e-Text Main Menu Textbook Table of Contents
10. 10. from which E  V0 1 lnb=a a DNa  V0 a lnb=a Q  V02aL a lnb=a C  2L lnb=a 17 which agrees with our results in Chap. 5. hExample 7.3 Now let us assume that V is a function only of in cylindrical coordinates. We might look at the physical problem first for a change and see that equipotential surfaces are given by  constant. These are radial planes. Boundary conditions might be V  0 at  0 and V  V0 at  , leading to the physical problem detailed in Fig. 7.1. Laplace's equation is now 1 2 @2 V @2  0 We exclude  0 and have d2 V d2  0 The solution is V  A  B 204 ENGINEERING ELECTROMAGNETICS FIGURE 7.1 Two infinite radial planes with an interior angle . An infinitesimal insulating gap exists at  0. The potential field may be found by applying Laplace's equation in cylindrical coordinates. | | | | v v e-Text Main Menu Textbook Table of Contents
11. 11. The boundary conditions determine A and B, and V  V0 18 Taking the gradient of (18) produces the electric field intensity, E  À V0a 19 and it is interesting to note that E is a function of and not of . This does not contradict our original assumptions, which were restrictions only on the potential field. Note, however, that the vector field E is a function of : A problem involving the capacitance of these two radial planes is included at the end of the chapter. hExample 7.4 We now turn to spherical coordinates, dispose immediately of variations with respect to only as having just been solved, and treat first V  Vr: The details are left for a problem later, but the final potential field is given by V  V0 1 r À 1 b 1 a À 1 b 20 where the boundary conditions are evidently V  0 at r  b and V  V0 at r  a, b a. The problem is that of concentric spheres. The capacitance was found previously in Sec. 5.10 (by a somewhat different method) and is C  4 1 a À 1 b 21 hExample 7.5 In spherical coordinates we now restrict the potential function to V  V, obtaining 1 r2 sin d d sin dV d  0 We exclude r  0 and  0 or and have POISSON'S AND LAPLACE'S EQUATIONS 205 | | | | v v e-Text Main Menu Textbook Table of Contents
12. 12. sin dV d  A The second integral is then V   A d sin  B which is not as obvious as the previous ones. From integral tables (or a good memory) we have V  A ln tan 2  B The equipotential surfaces are cones. Fig. 7.2 illustrates the case where V  0 at  =2 and V  V0 at  , =2: We obtain V  V0 ln tan 2 ln tan 2 22 In order to find the capacitance between a conducting cone with its vertex separated from a conducting plane by an infinitesimal insulating gap and its axis normal to the plane, let us first find the field strength: E  ÀrV  1 r @V @ a  À V0 r sin ln tan 2 a The surface charge density on the cone is then 206 ENGINEERING ELECTROMAGNETICS FIGURE 7.2 For the cone  at V0 and the plane  =2 at V  0, the potential field is given by V  V0lntan =2=lntan =2: | | | | v v e-Text Main Menu Textbook Table of Contents
14. 14. simultaneously. Such an ordeal is beyond the scope of this text, and we shall therefore assume a reasonably large amount of information. As an example, let us select a pn junction between two halves of a semi- conductor bar extending in the x direction. We shall assume that the region for x 0 is doped p type and that the region for x 0 is n type. The degree of doping is identical on each side of the junction. To review qualitatively some of the facts about the semiconductor junction, we note that initially there are excess holes to the left of the junction and excess electrons to the right. Each diffuses across the junction until an electric field is built up in such a direction that the diffusion current drops to zero. Thus, to prevent more holes from moving to the right, the electric field in the neighborhood of the junction must be directed to the left; Ex is negative there. This field must be produced by a net positive charge to the right of the junction and a net negative charge to the left. Note that the layer of positive charge consists of two partsÐthe holes which have crossed the junction and the positive donor ions from which the electrons have departed. The negative layer of charge is constituted in the opposite manner by electrons and negative acceptor ions. The type of charge distribution which results is shown in Fig. 7:3a, and the negative field which it produces is shown in Fig. 7:3b. After looking at these two figures, one might profitably read the previous paragraph again. A charge distribution of this form may be approximated by many different expressions. One of the simpler expressions is v  2v0 sech x a tanh x a 24 which has a maximum charge density v;max  v0 that occurs at x  0:881a. The maximum charge density v0 is related to the acceptor and donor concentrations Na and Nd by noting that all the donor and acceptor ions in this region (the depletion layer) have been stripped of an electron or a hole, and thus v0  eNa  eNd Let us now solve Poisson's equation, r2 V  À v subject to the charge distribution assumed above, d2 V dx2  À 2v0 sech x a tanh x a in this one-dimensional problem in which variations with y and z are not present. We integrate once, dV dx  2v0a sech x a  C1 and obtain the electric field intensity, 208 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
15. 15. POISSON'S AND LAPLACE'S EQUATIONS 209 FIGURE 7.3 a The charge density, b the electric field intensity, and c the potential are plotted for a pn junction as functions of distance from the center of the junction. The p-type material is on the left, and the n-type is on the right. | | | | v v e-Text Main Menu Textbook Table of Contents
16. 16. Ex  À 2v0a sech x a À C1 To evaluate the constant of integration C1, we note that no net charge density and no fields can exist far from the junction. Thus, as x 3 ÆI, Ex must approach zero. Therefore C1  0, and Ex  À 2v0a sech x a 25 Integrating again, V  4v0a2 tanÀ1 ex=a  C2 Let us arbitrarily select our zero reference of potential at the center of the junc- tion, x  0, 0  4v0a2 4  C2 and finally, V  4v0a2 tanÀ1 ex=a À 4 26 Fig. 7.3 shows the charge distribution (a), electric field intensity (b), and the potential (c), as given by (24), (25), and (26), respectively. The potential is constant once we are a distance of about 4a or 5a from the junction. The total potential difference V0 across the junction is obtained from (26), V0  Vx3I À Vx3ÀI  2v0a2 27 This expression suggests the possibility of determining the total charge on one side of the junction and then using (27) to find a junction capacitance. The total positive charge is Q  S I 0 2v0sech x a tanh x a dx  2v0aS where S is the area of the junction cross section. If we make use of (27) to eliminate the distance parameter a, the charge becomes Q  S  2v0V0 r 28 Since the total charge is a function of the potential difference, we have to be careful in defining a capacitance. Thinking in ``circuit'' terms for a moment, I  dQ dt  C dV0 dt 210 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
17. 17. and thus C  dQ dV0 By differentiating (28) we therefore have the capacitance, C   v0 2V0 r S  S 2a 29 The first form of (29) shows that the capacitance varies inversely as the square root of the voltage. That is, a higher voltage causes a greater separation of the charge layers and a smaller capacitance. The second form is interesting in that it indicates that we may think of the junction as a parallel-plate capacitor with a ``plate'' separation of 2a. In view of the dimensions of the region in which the charge is concentrated, this is a logical result. Poisson's equation enters into any problem involving volume charge den- sity. Besides semiconductor diode and transistor models, we find that vacuum tubes, magnetohydrodynamic energy conversion, and ion propulsion require its use in constructing satisfactory theories. D7.4. In the neighborhood of a certain semiconductor junction the volume charge density is given by v  750 sech 106 x tanh x C=m3 . The dielectric constant of the semiconductor material is 10 and the junction area is 2 Â 10À7 m2 . Find: a V0; b C; c E at the junction. Ans. 2.70 V; 8.85 pF; 2.70 MV/m D7.5. Given the volume charge density v  À2 Â 107 0  x p C=m3 in free space, let V  0 at x  0 and V  2 V at x  2:5 mm. At x  1 mm, find: a V; b Ex: Ans. 0.302 V; À555 V/m 7.5 PRODUCT SOLUTION OF LAPLACE'S EQUATION In this section we are confronted with the class of potential fields which vary with more than one of the three coordinates. Although our examples are taken in the cartesian coordinate system, the general method is applicable to the other coor- dinate systems. We shall avoid those applications, however, because the potential fields are given in terms of more advanced mathematical functions, such as Bessel functions and spherical and cylindrical harmonics, and our interest now does not lie with new mathematical functions but with the techniques and meth- ods of solving electrostatic field problems. We may give ourselves a general class of problems by specifying merely that the potential is a function of x and y alone, so that POISSON'S AND LAPLACE'S EQUATIONS 211 | | | | v v e-Text Main Menu Textbook Table of Contents
18. 18. @2 V @x2  @2 V @y2  0 30 We now assume that the potential is expressible as the product of a function of x alone and a function of y alone. It might seem that this prohibits too many solutions, such as V  x  y, or any sum of a function of x and a function of y, but we should realize that Laplace's equation is linear and the sum of any two solutions is also a solution. We could treat V  x  y as the sum of V1  x and V2  y, where each of these latter potentials is now a (trivial) product solution. Representing the function of x by X and the function of y by Y, we have V  X Y 31 which is substituted into (30), Y @2 X @x2  X @2 Y @y2  0 Since X does not involve y and Y does not involve x, ordinary derivatives may be used, Y d2 X dx2  X d2 Y dy2  0 32 Equation (32) may be solved by separating the variables through division by XY, giving 1 X d2 X dx2  1 Y d2 Y dy2  0 or 1 X d2 X dx2  À 1 Y d2 Y dy2 Now we need one of the cleverest arguments of mathematics: since 1=Xd2 X=dx2 involves no y and À1=Yd2 Y=dy2 involves no x, and since the two quantities are equal, then 1=Xd2 X=dx2 cannot be a function of x either, and similarly, À1=Yd2 Y=dy2 cannot be a function of y! In other words, we have shown that each of these terms must be a constant. For convenience, let us call this constant 2 ; 1 X d2 X dx2  2 33 À 1 Y d2 Y dy2  2 34 The constant 2 is called the separation constant, because its use results in separating one equation into two simpler equations. 212 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
19. 19. Equation (33) may be written as d2 X dx2  2 X 35 and must now be solved. There are several methods by which a solution may be obtained. The first method is experience, or recognition, which becomes more powerful with practice. We are just beginning and can barely recognize Laplace's equation itself. The second method might be that of direct integration, when applicable, of course. Applying it here, we should write d dX dx  2 X dx dX dx  2  X dx and then pass on to the next method, for X is some unknown function of x, and the method of integration is not applicable here. The third method we might describe as intuition, common sense, or inspection. It involves taking a good look at the equation, perhaps putting the operation into words. This method will work on (35) for some of us if we ask ourselves, ``What function has a second derivative which has the same form as the function itself, except for multiplica- tion by a constant?'' The answer is the exponential function, of course, and we could go on from here to construct the solution. Instead, let us work with those of us whose intuition is suffering from exposure and apply a very powerful but long method, the infinite-power-series substitution. We assume hopefully that X may be represented by X  I a0 anxn and substitute into (35), giving d2 X dx2  I 0 nn À 1anxnÀ2  2 I 0 anxn If these two different infinite series are to be equal for all x, they must be identical, and the coefficients of like powers of x may be equated term by term. Thus 2 Â 1 Â a2  2 a0 3 Â 2 Â a3  2 a1 and in general we have the recurrence relationship n  2n  1an2  2 an POISSON'S AND LAPLACE'S EQUATIONS 213 | | | | v v e-Text Main Menu Textbook Table of Contents
20. 20. The even coefficients may be expressed in terms of a0 as a2  2 1 Â 2 a0 a4  2 3 Â 4 a2  4 4! a0 a6  6 6! a0 and, in general, for n even, as an  n n! a0 n even For odd values of n, we have a3  2 2 Â 3 a1  3 3! a1 a5  5 5! a1 and in general, for n odd, an  n n! a1 n odd Substituting back into the original power series for X, we obtain X  a0 I 0;even n n! xn  a1 I 1;odd n n! xn or X  a0 I 0;even xn n!  a1 I 1;odd xn n! Although the sum of these two infinite series is the solution of the differential equation in x, the form of the solution may be improved immeasurably by recognizing the first series as the hyperbolic cosine, cosh x  I 0;even xn n!  1  x2 2!  x4 4!  . . . and the second series as the hyperbolic sine, sinh x  I 1;odd xn n!  x  x3 3!  x5 5!  : 214 ENGINEERING ELECTROMAGNETICS | | | | v v e-Text Main Menu Textbook Table of Contents
21. 21. The solution may therefore be written as X  a0 cosh x  a1 sinh x or X  A cosh x  B sinh x where the slightly simpler terms A and B have replaced a0 and a1=, respectively, and are the two constants which must be evaluated in terms of the boundary conditions. The separation constant is not an arbitrary constant as far as the solution of (35) is concerned, for it appears in that equation. An alternate form of the solution is obtained by expressing the hyperbolic functions in terms of exponentials, collecting terms, and selecting new arbitrary constants, AH and BH ; X  AH ex  BH eÀx Turning our attention now to (34), we see the solution proceeds along similar lines, leading to two power series representing the sine and cosine, and we have Y  C cos y  D sin y from which the potential is V  X Y  A cosh x  B sinh xC cos y  D sin y 36 Before describing a physical problem and forcing the constants appearing in (36) to fit the boundary conditions prescribed, let us consider the physical nature of the potential field given by a simple choice of these constants. Letting A  0, C  0, and BD  V1, we have V  V1 sinh x sin y 37 The sinh x factor is zero at x  0 and increases smoothly with x, soon becoming nearly exponential in form, since sinh x  1 2 ex À eÀx  The sin y term causes the potential to be zero at y  0, y  =, y  2=, and so forth. We therefore may place zero-potential conducting planes at x  0, y  0, and y  =. Finally, we can describe the V1 equipotential surface by setting V  V1 in (37), obtaining sinh x sin y  1 or y  sinÀ1 1 sinh x This is not a familiar equation, but a hand calculator or a set of tables can furnish enough material values to allow us to plot y as a function of x. Such a POISSON'S AND LAPLACE'S EQUATIONS 215 | | | | v v e-Text Main Menu Textbook Table of Contents
22. 22. curve is shown in Fig. 7.4. Note that the curve is double-valued and symmetrical about the line y  =2 when y is restricted to the interval between 0 and . The information of Fig. 7.4 is transferred directly to the V  0 and V  V1 equipo- tential conducting surfaces in Fig. 7.5. The surfaces are shown in cross section, since the potential is not a function of z. It is very unlikely that we shall ever be asked to find the potential field of these peculiarly shaped electrodes, but we should bear in mind the possibility of combining a number of the fields having the form given by (36) or (37) and thus satisfying the boundary conditions of a more practical problem. We close this chapter with such an example. The problem to be solved is that shown in Fig. 7.6. The boundary condi- tions shown are V  0 at x  0, y  0, and y  b, and V  V0 at x  d for all y 216 ENGINEERING ELECTROMAGNETICS FIGURE 7.4 A graph of the double-valued function y  sinÀ1 1= sinh x; 0 y : FIGURE 7.5 Cross section of the V  0 and V  V1 equipotential surfaces for the potential field V  V1 sinh x sin y: | | | | v v e-Text Main Menu Textbook Table of Contents
23. 23. between 0 and b. It is immediately apparent that the potential field given by(37) and outlined in Fig. 7.5 satisfies two of the four boundary conditions. A third condition, V  0 at y  b, may be satisfied by the choice of a, for the substitu- tion of these values of (37) leads to the equation 0  V1 sinh x sin b which may be satisfied by setting b  m m  1; 2; 3; . . . or  m b The potential function V  V1 sinh mx b sin my b 38 thus produces the correct potential at x  0, y  0, and y  b, regardless of the choice of m or the value of V1. It is impossible to choose m or V1 in such a way that V  V0 at x  d for each and every value of y between 0 and b. We must combine an infinite number of these fields, each with a different value of m and a corresponding value of V1; V  I m0 V1m sinh mx b sin my b The subscript on V1m indicates that this amplitude factor will have a different value for each different value of m. Applying the last boundary condition now, POISSON'S AND LAPLACE'S EQUATIONS 217 FIGURE 7.6 Potential problem requiring an infinite summation of fields of the form V  V1 sinh x sin y. A similar configuration was analyzed by the iteration method in Chap. 6. | | | | v v e-Text Main Menu Textbook Table of Contents
24. 24. V0  I m0 V1m sinh md d sin my b 0 y b; m  1; 2; . . . Since V1m sinh md=b is a function only of m, we may simplify the expression by replacing this factor by cm; V0  I m0 cm sin my b 0 y b; m  1; 2; . . . This is a Fourier sine series, and the cm coefficients may be determined by the standard Fourier-series methods1 if we can interpret V0 as a periodic function of y. Since our physical problem is bounded by conducting planes at y  0 and y  b, and our interest in the potential does not extend outside of this region, we may define the potential at x  d for y outside of the range 0 to b in any manner we choose. Probably the simplest periodic expression is obtained by selecting the interval 0 y b as the half-period and choosing V  ÀV0 in the adjacent half- period, or V  V0 x  d; 0 y b V  ÀV0 x  d; b y 2b The cm coefficients are then cm  1 b b 0 V0 sin my b dy  2b b ÀV0 sin my b dy ! leading to cm  4V0 m m odd  0 m even However, cm  V1m sinh md=b, and therefore V1m  4V0 m sinh md=b m odd only which may be substituted into (38) to give the desired potential function, V  4V0 I 1;odd 1 m sinh mx=b sinh md=b sin my b 39 The map of this field may be obtained by evaluating (39) at a number of points and drawing equipotentials by interpolation between these points. If we let b  d and V0  100, the problem is identical with that used as the example in 218 ENGINEERING ELECTROMAGNETICS 1 Fourier series are discussed in almost every electrical engineering text on circuit theory. The authors are partial to the Hayt and Kemmerly reference given in the Suggested References at the end of the chapter. | | | | v v e-Text Main Menu Textbook Table of Contents
25. 25. the discussion of the iteration method. Checking one of the grid points in that problem, we let x  d=4  b=4, y  b=2  d=2, and V0  100 and obtain V  400 I 1;odd 1 m sinh m=4 sinh m sin m 2  400 sinh =4 sinh À 1 3 sinh 3=4 sinh 3  1 5 sinh 5=4 sinh 5 À . . .  400 0:8687 11:549 À 5:228 3 Â 6195:8  . . .  9:577 À 0:036  . . .  9:541 V The equipotentials are drawn for increments of 10 V in Fig. 7.7, and flux lines have been added graphically to produce a curvilinear map. The material covered in this discussion of the product solution was more difficult than much of the preceding work, and moreover, it presented three new ideas. The first new technique was the assumption that the potential might be expressed as the product of a function of x and a function of y, and the resultant separation of Laplace's equation into two simpler ordinary differential equa- tions. The second new approach was employed when an infinite-power-series solution was assumed as the solution for one of the ordinary differential equa- tions. Finally, we considered an example which required the combination of an infinite number of simpler product solutions, each having a different amplitude and a different variation in one of the coordinate directions. All these techniques POISSON'S AND LAPLACE'S EQUATIONS 219 FIGURE 7.7 The field map corresponding to V  4V0 I 1;odd 1 m sinhmx=b sinhmd=b sin my b with b  d and V0  100 V. | | | | v v e-Text Main Menu Textbook Table of Contents